## Related questions with answers

At its peak, a tornado is 60.0 m in diameter and carries 500 km/h winds. What is its angular velocity in revolutions per second?

Solution

VerifiedFirst, we convert the linear speed units to meters per second. Then, we proceed to calculate the angular velocity with the given diameter $\left(d\right)$. Note that we have substituted the radius $\left(r\right)$ for $d/2$.

$\begin{align*} v&=500 \text{ } \dfrac{\text{km}}{\text{h}} \cdot \dfrac{1000 \text{ m}}{1 \text{ km}} \cdot \dfrac{1 \text{ h}}{3600\text{ s}} \\ &=139 \text{ } \dfrac{\text{m}}{\text{s}} \\ \omega&=\dfrac{v}{r} \\ &=\dfrac{2v}{d} \\ &=\dfrac{2\left(139\right)}{60.0} \\ &=4.63 \text{ } \dfrac{\text{rad}}{\text{s}} \\ &=4.63 \text{ } \dfrac{\text{rad}}{\text{s}} \cdot \dfrac{1 \text{ rev}}{2\pi \text{ rad}} \\ &=0.737 \text{ } \dfrac{\text{rev}}{\text{s}} \end{align*}$

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